In this blog posting we’ll take a look at the dynamic performance of a stepper motor.
We’ve established the fact that there has to be an error between the rotor’s position and the stator’s position in order for the step motor to generate any torque.
This torque can be generated by forcing the rotor out of its stable position by twisting the shaft or by causing the motor to step to a new position. The stepping action is accomplished by changing the current in the stator’s windings in an appropriate sequence that causes the rotor to turn CW or CCW. And we’ve covered all of that in previous blog postings, so if this is all new to you, then I suggest you back up a bunch of postings and start from the beginning.
We also know that it takes time, even though it’s in the microseconds, to get the current to change in the winding. Now start taking faster and faster steps and the drive is going to reach a point where the motor is going to be stepping faster than the drive’s ability to get the full current into the winding.
Let’s assume our micro stepping drive is operating at 256 uSteps per full steps, or 51,200 steps per rev and we’re spinning this motor at a comfortable 25 revs per second (1, 500 RPM.) 51,200 steps/rev * 25 revs/second =1,280,000 uSteps per second.
The time period of one uStep is 1/1,280,000 = 0.781 micro seconds
Much faster than the time it takes to pump the current up to its full rating.
So what happens? The torque begins to fall off as the speed increases.
See figure one below:
With the drive powered with 12 VDC the motor produces its full torque only out to about 800 full steps/second (FS/S) and then the torque begins to fall off as the speed increases.
Bump the power supply voltage up to 48 VDC on the same motor and drive and the performance is better. The torque is flat all the way to about 1,800 FS/S and again falls off as the speed increases. Note that the toque that’s available at 5,000 FS/S is significantly higher with the drive that’s powered with 48 VDC than with the one powered with 12 VDC. Again, that’s because the current can get into the winding faster, so the torque doesn’t fall off as fast.
Increase the power supply voltage up to 75 VDC and again you get significant increases in torque at higher speeds.
Now with a real motor and drive the curves won’t be “so perfect” as the one drawn here, but you get the idea.
So now apply this knowledge to moving a load.
I’m just going to pick a torque requirement out of the air (50 oz-in) instead of calculating the inertia, friction load, acceleration and deceleration rates, or even take into consideration whether the load is moving horizontal or vertical. These and other parameters have an affect on the application and of course on the torque requirement.
If our speed requirement is for only 1,000 FS/S then our 100 oz-in motor and a 12 VDC drive will work, but if our speed requirement is to 2,500 FS/S then the 12 VDC solution is marginal. More so than you might think, because we really should have a torque safety factor of about 50%.
Choosing the 48 VDC drive and we’re back in business.
The significance of this exercise is to show that the speed-torque curve is a very useful tool and the motor-drive choice should always have the application’s torque-speed operating point below the motor’s performance curve.
If the motor’s torque requirement goes above the curve, what happens?
The motor stalls and position is lost. The motor can’t restart at high frequencies (typically above 500 to1,000 FS/S.) The control would have to cease sending steps for a period of time to allow the motor to find its stable rest position and then ramp the motor back up to its high speed.
A 50% safety factor takes into account that most speed-torque curves are “typical” performance curves. That means that some motors might perform better than the curves and some might perform worse. Plus you should have some safety factor built in accommodate any acceleration/deceleration perturbations.
Next: Is there a magic elixir that prevents the motor from stalling?